# Thread: Cylindrical Coordinates and vectors

1. ## Cylindrical Coordinates and vectors

i am not sure what to do because this problem has two origins which is not discussed in my electromagnetics book:
cyl coordiinates (rho,phi,z)
Given:
$\displaystyle \vec{A} = 3\vec{a}_{\rho A} + 5\vec{a}_{\phi A} + 2\vec{a}_{zA}$ at point (5, 30 degrees, 7)

$\displaystyle \vec{B} = 5\vec{a}_{\rho B} + 5\vec{a}_{\phi B} + 2\vec{a}_{zB}$ at point (2, 45 degrees, 7)

Find:
$\displaystyle \vec{A} dot \vec{B}$
$\displaystyle \vec{A} + \vec{B}$
$\displaystyle \vec{A} X \vec{B}$

can you do at least one, (hardest which is cross product)
thanks

i am not sure what to do because this problem has two origins which is not discussed in my electromagnetics book:
cyl coordiinates (rho,phi,z)
Given:
$\displaystyle \vec{A} = 3\vec{a}_{\rho A} + 5\vec{a}_{\phi A} + 2\vec{a}_{zA}$ at point (5, 30 degrees, 7)

$\displaystyle \vec{B} = 5\vec{a}_{\rho B} + 5\vec{a}_{\phi B} + 2\vec{a}_{zB}$ at point (2, 45 degrees, 7)

Find:
$\displaystyle \vec{A} dot \vec{B}$
$\displaystyle \vec{A} + \vec{B}$
$\displaystyle \vec{A} X \vec{B}$

can you do at least one, (hardest which is cross product)
thanks
The determinant representation of the cross product also works with $\displaystyle \rho$, $\displaystyle \phi$ and z unit vectors ......

3. shall i instantly cross product this ignoring its origins?

shall i instantly cross product this ignoring its origins?
Why not? The result will be a vector in cylindrical coordinates.

i am not sure what to do because this problem has two origins which is not discussed in my electromagnetics book:
cyl coordiinates (rho,phi,z)
Given:
$\displaystyle \vec{A} = 3\vec{a}_{\rho A} + 5\vec{a}_{\phi A} + 2\vec{a}_{zA}$ at point (5, 30 degrees, 7)

$\displaystyle \vec{B} = 5\vec{a}_{\rho B} + 5\vec{a}_{\phi B} + 2\vec{a}_{zB}$ at point (2, 45 degrees, 7)

Find:
$\displaystyle \vec{A} dot \vec{B}$
$\displaystyle \vec{A} + \vec{B}$
$\displaystyle \vec{A} X \vec{B}$

can you do at least one, (hardest which is cross product)
thanks
Worst comes to worst you can convert them to Cartesian coordinates, do the product, then convert back.

-Dan